Ron's Roots

(1+α)(1+β)=1+(α+β)+αβ. According to Vieta's Formulas: If α and β are the roots of ax^2+bx+c=0,then α+β= - b/a and αβ= c/a. So α+β= - b/a = -4/3 and αβ= c/a = 9/3=3. Hence (1+α)(1+β)=1+(α+β)+αβ=1+(-4/3)+3=8/3. 8/3 is the form of a/b. Therefore a+b=8+3=11.

We know that quadratic equations are of the form ax^2 + bx +c=0 where sum of roots = - ( \frac {b}{a} ) and product of roots = \frac {c}{a}. Here, sum of roots = (alpha + beta) = - ( \frac {4}{3} ) and product of roots = ( alpha * beta ) = \frac {9}{3} On expanding (1 + aplha)(1 + beta) we get 1 + ( alpha + beta) + (alpha * beta) . Putting values in this equation from above, we get 1 - ( \frac {4}{3} ) + ( \frac {9}{3} ) . Solve it to get /frac {8}{3} which is 11 in terms of a + b.

(1+α)(1+β)=1+(α+β)+αβ=1-4/3+9/3=8/3 a=8 , b=3 , a+b=3

We know that the sum of the roots is -4/3 and their product 3. So (1+a)(1+b)= 1+a*b +a+b =4-4/3=8/3.

Let S = $\alpha$ + $\beta$ and P = $\alpha$ $\beta$

From Vieta's Formulae for sum and product of roots we obtain S = - $\frac{4}{3}$ and P = $\frac{9}{3}$ = 3

Let X = (1 + $\alpha$ ) (1 + $\beta$ )

= 1 + $\alpha$ + $\beta$ + $\alpha$ $\beta$

= 1 + S + P

= 1 - $\frac{4}{3}$ + 3

= $\frac{8}{3}$ = $\frac{a}{b}$

Since $a$ and $b$ are co-prime positive integers, comparing from the final result we obtain $a$ + $b$ = 8 + 3 = 11 which is our answer.

Expanding $(1 + \alpha)(1 + \beta)$ , we get $1 + \alpha + \beta + \alpha\beta$ . In the given polynomial $3x^2 + 4x + 9$ , $a = 3, b = 4$ and $c = 9$ . According to Vieta's Formula, $\alpha + \beta = \frac{-b}{a} = -\frac{4}{3}$ and $\alpha\beta = \frac{c}{a} = \frac{9}{3} = 3$ .

Hence, $1 + \alpha + \beta + \alpha\beta = 1 + \left(\frac{-4}{3}\right) + 3 = \frac{8}{3}$ .

Since $8$ and $3$ are coprime, the answer is simply $8 + 3 = 11$ .

(1+α)(1+β)=1+α+β+αβ By Viete's Theorem ,we have: α+β= -4/3 αβ =9/3 So (1+α)(1+β)=1+(-4/3)+9/3=8/3 a+b=8+3 =11

For any given quadratic equation $ax^2 + bx + c$ , the roots are

$\alpha$ = $\frac{-b + \sqrt{b^2-4ac}}{2a}$

$\beta$ = $\frac{-b - \sqrt{b^2-4ac}}{2a}$

Hence

$\alpha$ + $\beta$ = $\frac{-b + \sqrt{b^2-4ac}}{2a}$ + $\frac{-b - \sqrt{b^2-4ac}}{2a}$ = $\frac{-b + \sqrt{b^2-4ac} + (-b) - \sqrt{b^2-4ac}}{2a}$ = $\frac{-2b}{2a}$ = $-\frac{b}{a}$

$\alpha\ \cdot \beta$ = $\frac{-b + \sqrt{b^2-4ac}}{2a} \cdot \frac{-b - \sqrt{b^2-4ac}}{2a}$ = $\frac{(-b + \sqrt{b^2-4ac}) \cdot (-b - \sqrt{b^2-4ac})}{4a^2}$ = $\frac{(-b)^2 - (\sqrt{b^2-4ac})^2}{4a^2}$ = $\frac{b^2 - b^2 + 4ac}{4a^2}$ = $\frac{4ac}{4a^2}$ = $\frac{c}{a}$

In the question, the formula is $3x^2 + 4x + 9$ .

Thus

$(1+\alpha) \cdot (1+\beta)$ = $1 + \alpha + \beta + \alpha \cdot \beta$ = $1 -\frac{4}{3} + \frac{9}{3}$ = $\frac{8}{3}$

8 and 3 are coprime, hence the answer is $8+3$ = $11$

Given: $3x^2+4x+9=0$

We have Sreedhara's formula, $x= \frac { -b \pm \sqrt{b^2-4ac}}{2a}$

i.e. $x= \frac { -4 \pm \sqrt {4^2-4*3*9}}{2*3}$ $= \frac { -4 \pm \sqrt {-92}}{6}$ $= \frac { -2 \pm i\sqrt {23}}{3}$

Now , let $\alpha = \frac { -2 + i\sqrt {23}}{3}$ and $\beta = \frac { -2 - i\sqrt {23}}{3}$

Thus, $(1+\alpha) = \frac { -2 + i\sqrt {23}}{3} +1$ $= \frac { 1 + i\sqrt {23}}{3}$

and $(1+\beta ) = \frac { -2 - i\sqrt {23}}{3} +1$ $= \frac { 1 - i\sqrt {23}}{3}$

Therefore, $(1+\alpha) (1+\beta) = (\frac{1+i\sqrt{23}}{3}) (\frac{1-i\sqrt{23}}{3})$ $(1+\alpha) (1+\beta) = (\frac{1^2 - i^2 (\sqrt{23})^2}{3*3}) = \frac{1+23}{9} =\frac{24}{9}=\frac{8}{3}=\frac{a}{b}$

therefore , a = 8 and b=3 and

a+b = 8+3 = 11.

$f(x)=3(x-\alpha)(x-\beta)=3x^2+4x+9$

$\Rightarrow f(-1)=3(1+\alpha)(1+\beta)=8 \Rightarrow (1+\alpha)(1+\beta)=\frac{8}{3}$

$\Rightarrow a=8,b=3 \Rightarrow a+b=\boxed{11}$